# Example of normal distribution in an intervalNormal Distribution Example

### Problem Data

It is required to calculate the **probability** of **P(485000<X<530000)** (interval) with the following data:

- Mean (μ)
- 500000
- Standard Deviation (σ)
- 15000
- Description
**Example of normal distribution in an interval**A customer has an investment portfolio whose mean value is $500,000 and whose standard deviation is $15,000. You have asked him to calculate the probability that the value of his portfolio is between $485,000 and $530,000.

## Solution

Below are the detailed calculations and graphs to obtain the **probability** according to the data provided:

### Step 1:

We have the normal random variable X with mean μ and variance σ^{2}; that is, ** X ~ N(μ,σ^{2})**, which will become the standard normal random variable,

**. To do so, we will apply the following transformation:**

*Z ~ N(0,1)*We have μ = 500000 and σ = 15000, then:

The exact result of the expression obtained is:

To check the result using the lower-tailed normal distribution (Z) table, we will make use of the following property:

In our problem it would be:

**1. For P(Z<2)**

The exact result of the expression obtained is:

To check the result we will use the left-tailed normal distribution (Z) table.

We will use our table to find the probability of ** P(Z<2.00)** as follows:

- In the rows of the table we place the value of
**2.0** - In the columns of the table we place the value of
**0.00** - The intersection of the indicated row and column will give us the value of
(see highlighted value in the table below).*P(Z<2.00)=0.9772*

**2. For P(Z<-1)**

The exact result of the expression obtained is:

To check the result we will use the left-tailed normal distribution (Z) table.

Since our table does not show negative values, we will use the symmetry of the normal distribution to solve the problem:

We will use our table to find the probability of ** P(Z<1.00)** as follows:

- In the rows of the table we place the value of
**1.0** - In the columns of the table we place the value of
**0.00** - The intersection of the indicated row and column will give us the value of
(see highlighted value in the table below).*P(Z<1.00)=0.8413*

Finally, we replace the value in the obtained expression:

Returning to the original expression, we replace the values obtained:

The probability that the random variable X lies in the interval [485000,530000] is **0.8186**

**Note:** The results obtained by the calculator and the one obtained by the table may have small differences due to rounding issues.

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